Question
If $p(y) = y^2 – 3√2 + 1,$ then find $p( 3√2 ).$
✓
Answer
$p(y) = y^2 – 3√2 y + 1$
Putp= $3√2$ in the given polynomial.
$\therefore p( 3√2 ) = (3√2 )^2 – 3√2 (3√2 ) + 1$
$= 9 x 2 – 9 x 2 + 1$
$= 18 – 18 + 1$
$\therefore p( 3√2 ) = 1$
Putp= $3√2$ in the given polynomial.
$\therefore p( 3√2 ) = (3√2 )^2 – 3√2 (3√2 ) + 1$
$= 9 x 2 – 9 x 2 + 1$
$= 18 – 18 + 1$
$\therefore p( 3√2 ) = 1$
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